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Zinaida [17]
3 years ago
10

How many moles of air must escape from a 10-m  8.0-m  5.0-m room when the temperature is raised from 0c to 20c? assume the p

ressure remains unchanged at one atmosphere while the room is heated?
Chemistry
1 answer:
slega [8]3 years ago
3 0
Data:

p = 1 atm
V = 10 m * 8 m * 5 m = 400 m^3 = 400,000 liter

To = 0 + 273.15K = 273.15K
Tf = 20 + 273.15K = 293.15K

No - Nf =?

2) Formula

pV = NRT => N = pV / (RT)

3) solution

No = pV / (RTo)

Nf = pV / (RTf)

=> No - Nf = [pv / R] [ 1 / To - 1 / Tf ]

=> No - Nf = [1atm*400,000liter / 0.0821 atm*liter/K*mol ] [ 1 / 273.15 - 1 / 293.15]

No - Nf = 1216.9 moles ≈ 1217 moles

Answer: 1217 moles
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At 25.0°c, a solution has a concentration of 3.179 m and a density of 1.260 g/ml. the density of the solution at 50.0°c is 1.249
oksano4ka [1.4K]

Answer: -

3.151 M

Explanation: -

Let the volume of the solution be 1000 mL.

At 25.0 °C, Density = 1.260 g/ mL

Mass of the solution = Density x volume

= 1.260 g / mL x 1000 mL

= 1260 g

At 25.0 °C, the molarity = 3.179 M

Number of moles present per 1000 mL = 3.179 mol

Strength of the solution in g / mol

= 1260 g / 3.179 mol = 396.35 g / mol (at 25.0 °C)

Now at 50.0 °C

The density is 1.249 g/ mL

Mass of the solution = density x volume = 1.249 g / mL x 1000 mL

= 1249 g.

Number of moles present in 1249 g = Mass of the solution / Strength in g /mol

= \frac{1249 g}{396.35 g/mol}

= 3.151 moles.

So 3.151 moles is present in 1000 mL at 50.0 °C

Molarity at 50.0 °C = 3.151 M

7 0
3 years ago
How many protons do calcium have
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